Methanol (ch3oh) can be produced by the following reaction: co(g) 1 2h2(g) 88n ch3oh(g) hydrogen at stp flows into a reactor at a rate of 16.0 l/min. carbon monoxide at stp flows into the reactor at a rate of 25.0 l/min. if 5.30 g of methanol is produced per minute, what is the percent yield of the reaction?

Explanation:

It is know that, 1 mole of a gas at STP occupies a volume of 22. 4 L. Hence, calculate the number of moles of CO as follows.

     No. of moles of CO being fed per minute = per minute

                  = 1.116 moles per minute

No. of moles of being fed per minute = per minute

                 = 0.714 moles per minute

As, the balanced reaction equation is as follows.

Here, is the limiting reagent.  Since, 2 moles is equivalent to 1 mol methanol produced .

Or,          0.71 moles of contains

                            = 0.357 moles of methanol.

Therefore, theoretical yield of methanol = 0.357 moles per minute.

Observed yield of methanol = 5.30 gram per minute

                                         =

                                          = 0.165 moles per minute

Nowe, we will calculate the percentage yield as follows.

       % yield =

                      =

                      = 46.32%

Thus, we can conclude that percent yield of the reaction is 46.32%.

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